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PtichkaEL [24]
3 years ago
15

What is the total current through the circuit? 6 ohms 9V 9 ohms 12 ohms​

Physics
1 answer:
Goshia [24]3 years ago
7 0

Answer:

Here is no any circuit

but ( hint)

Kirchoff's current law is used here

Kirchhoff's current law (1st Law) states that the current flowing into a node (or a junction) must be equal to the current flowing out of it. This is a consequence of charge conservation.

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The relationship between an object's acceleration and its mass
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Answer:

Newton's second law

Explanation:

The relationship between mass and acceleration is described in Newton's Second Law of Motion. His Second Law states that the more mass an object has, more force is necessary for it to accelerate.

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4 years ago
Interchangeably means​
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Interchangeable is something that can be used in the same manner as something else without any important differences.

or in a more simple way

something that can be exchange

For example when two sweaters look almost the same and you could easily wear either one with a particular skirt, this is an example of when the sweaters can be worn interchangeably.

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3 years ago
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A soccer player practices kicking the ball into the goal from halfway down the soccer field. The time it takes for the ball to g
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This is correct, I just did the test. Yes, displacement is 45 meters, elapsed time is three seconds, and the direction is toward the goal.

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3 years ago
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What were some major reasons for the american television from analog broadcast to digital broadcast?
Vika [28.1K]

Answer: It frees up valuable portions of the broadcast spectrum, it has better audio and picture quality, and there are more options on digital broadcasting

Explanation:

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4 years ago
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2. A 20 cm object is placed 10cm in front of a convex lens of focal length 5cm. Calculate
adoni [48]

Answer:

<u> </u><u>»</u><u> </u><u>Image</u><u> </u><u>distance</u><u> </u><u>:</u>

{ \tt{ \frac{1}{v}  +  \frac{1}{u} =  \frac{1}{f}  }} \\

  • v is image distance
  • u is object distance, u is 10 cm
  • f is focal length, f is 5 cm

{ \tt{ \frac{1}{v} +  \frac{1}{10} =  \frac{1}{5}   }} \\  \\  { \tt{ \frac{1}{v}  =  \frac{1}{10} }} \\  \\ { \tt{v = 10}} \\  \\ { \underline{ \underline{ \pmb{ \red{ \: image \: distance \: is \: 10 \: cm \:  \: }}}}}

<u> </u><u>»</u><u> </u><u>Magnification</u><u> </u><u>:</u>

• Let's derive this formula from the lens formula:

{ \tt{ \frac{1}{v}  +  \frac{1}{u} =  \frac{1}{f}  }} \\

» Multiply throughout by fv

{ \tt{fv( \frac{1}{v} +  \frac{1}{u} ) = fv( \frac{1}{f}  )}} \\   \\ { \tt{ \frac{fv}{v}  +  \frac{fv}{u}  =  \frac{fv}{f} }} \\  \\  { \tt{f + f( \frac{v}{u} ) = v}}

• But we know that, v/u is M

{ \tt{f + fM = v}} \\  { \tt{f(1 +M) = v }} \\ { \tt{1 +M =  \frac{v}{f}  }} \\  \\ { \boxed{ \mathfrak{formular :  } \: { \tt{ M =  \frac{v}{f}  - 1 }}}}

  • v is image distance, v is 10 cm
  • f is focal length, f is 5 cm
  • M is magnification.

{ \tt{M =  \frac{10}{5} - 1 }} \\  \\ { \tt{M = 5 - 1}} \\  \\ { \underline{ \underline{ \pmb{ \red{ \: magnification \: is \: 4}}}}}

<u> </u><u>»</u><u> </u><u>Nature</u><u> </u><u>of</u><u> </u><u>Image</u><u> </u><u>:</u>

  • Image is magnified
  • Image is erect or upright
  • Image is inverted
  • Image distance is identical to object distance.
4 0
2 years ago
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